Today I learned this interesting fact from Jerry Kaminker: If $A$ is an abelian group such that $\mathrm{Hom}(A,\mathbb{Z})$ and $\mathrm{Ext}(A,\mathbb{Z})$ are both countably generated, then in fact $A$ is finitely generated. This is known in the literature, in some old papers by Nunke-Rotman, Chase, and Mitchell. It makes me interested in possible generalizations.

Suppose that $M$ is a left module over a ring $R$ and that $\mathrm{Ext}^k(M,R)$ is countably generated for all $k$. For which $R$ can you conclude that $M$ is finitely generated or, better, finitely resolved? Any commutative Noetherian ring with finite projective dimension? Is there a countability restriction missing from this proposed generalization? What about non-commutative rings?

The result has been stated for any countable PID rather than just for $\mathbb{Z}$. In fact Mitchell says that if $R$ is a countable PID and $M$ is infinitely generated, then $$|\mathrm{Hom}(M,R)|\cdot|\mathrm{Ext}(M,R)| = 2^{|M|}.$$

Victor in the comments asks for an application for this result for abelian groups. The original motivation was to extract information about what is possible for the cohomology of a topological space. The homology can be anything, provided that $H_0(X)$ is free and non-trivial. There are various obvious and not-so-obvious impossible choices for cohomology. For instance, this result implies that $H^*(X)$ cannot be countably infinitely generated as an abelian group. (And if it is so as a ring, then the degrees of the generators have to go to $\infty$.) I suspect that Jerry needs it for a similar reason.

I found the result interesting because it gives an "external" criterion for whether a countable abelian group is finitely generated. Even though the forgetful functor to Set does not distinguish countable groups, it does sometimes distinguish their dual or derived dual groups. One of the things that it can determine is whether the original group was finitely generated. Also, assuming that the result holds for all PIDs, it is a natural and non-trivial generalization of the fact that the algebraic dual of an infinite-dimensional vector space $V$ satisfies $$\dim V^* = 2^{\dim V}.$$ Everyone learns this for vector spaces, so it's cool to have a module version.

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    $\begingroup$ For the case of abelian groups, does this characterization have any interesting applications? $\endgroup$ Jul 23, 2010 at 7:56
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    $\begingroup$ In addition to the original sources, a textbook exposition of the abelian group case can be found in Proposition 3F.12 of my algebraic topology book. (Don't you just hate it when authors are constantly referring to their own books?) $\endgroup$ Jul 23, 2010 at 16:11
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    $\begingroup$ @Allen: I certainly don't hate it if it's your book! $\endgroup$ Jul 23, 2010 at 16:21

1 Answer 1


This is an answer to one of the questions asked:

It is not sufficient for $R$ to be a commutative noetherian ring with finite global dimension.

Let $R=\mathbb{Z}_p$, the $p$-adic integers, which is a principal ideal domain, and so has global dimension $1$. It is well known that $\operatorname{Ext}^1_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p)=0$, and clearly $\operatorname{Hom}_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p)=0$.

So for $M=\mathbb{Q}_p$, $\operatorname{Ext}_R^k(M,R)=0$ (so certainly countably generated!) for all $k\geq0$, but $M$ is not finitely generated as an $R$-module.

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    $\begingroup$ Although it is well known that $\operatorname{Ext}^1_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Z}_p)=0$, and it's not too hard to prove, I couldn't find a convenient reference giving a proof. Does anybody know one? $\endgroup$ Aug 25, 2020 at 12:00
  • $\begingroup$ I just noticed this a long time later. Thanks! $\endgroup$ Apr 19 at 14:52

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